1. Modulation and OFDM

2. Communication Exchange of information from point A to point B
Transmit
Receive

3. Wireless Communication
Exchange of information from point A to point B
Transmit
Receive

4. Wireless Communication
Exchange of information from point A to point B
Modulation
Upconvert
Downconvert
Demodulation

5. Modulation Converting bits to signals
These signals are later sent over the air
The receiver picks these signals and decodes transmitted data
Modulation
Signals (voltages)

6. Amplitude Modulation
Suppose we have 4 voltage levels (analog) to represent bits. 00 01 10 11
𝑉 4
- 𝑉 4
3𝑉 4
− 3𝑉 4

7. Amplitude Modulation - 𝑉 4 - 𝑉 4 00
3𝑉 4
𝑉 4 01
- 𝑉 4 10
3𝑉 4 11
− 3𝑉 4
𝑉 4
- 𝑉 4
Individual voltage levels are called as symbols
− 3𝑉 4

8. Modulated symbols for transmission
FFT -F
Frequency F

9. Received symbols with distortions
FFT -F
Frequency F

10. Demodulation
Tx bits 00
3𝑉 4
𝑉 4 01
- 𝑉 4 10
3𝑉 4 11
− 3𝑉 4
𝑉 4
- 𝑉 4
− 3𝑉 4
0 0
0 0
1 0
Rx bits decoded
1 0

11. Coping up with demodulation errors
If the noise is too high, there may be too many bit flips
Symbols for modulation to be chosen as a function of this noise
For example, if we want to eliminate bit flips completely, we can choose voltage levels as follows

12. Modulation with sparser symbols1 𝑉 −𝑉 𝑉 −𝑉

13. Received symbols with distortion1 𝑉 −𝑉 𝑉 −𝑉

14. Demodulation 1 𝑉 −𝑉 𝑉 −𝑉

15. That eliminated all the bit flips, which is good
However, what is the disadvantage of choosing only two voltage levels?
Takes longer to transmit, hence bit rate is very low

16. Bit rates
Bit per symbol 𝑛𝑏 𝑠
𝐵 (bandwidth)
Symbol rate (Baud rate) R s = 1 𝑡 𝑠
Frequency
Symbol duration 𝑡 𝑠 -F F 𝑉
FFT
𝑉 2
Bit rate 𝑅 𝑏 =𝑛 𝑏 𝑠 ∗ 𝑅 𝑠
- 𝑉 2
Symbol rate (Baud rate) R b =2∗𝐵 (bandwidth)
Bit rate 𝑅 𝑏 =2∗𝑛 𝑏 𝑠 ∗𝐵 −𝑉

17. Transmission of modulated symbols
The modulated message has zero center frequency (baseband)
Impractical to have antennas at that frequencies
Causes interference if everyone wants to use baseband ..
𝐵 (bandwidth)
𝐹 𝑐 =2.4𝐺𝐻𝑧
𝐹 𝑐 =−2.4𝐺𝐻𝑧
Upconversion
𝐵 (bandwidth)
𝑐𝑒𝑛𝑡𝑒𝑟 𝑓𝑟𝑒𝑞 𝐹 𝑐 =0

18. Upconversion  shifting center frequency𝑓 −𝑓
cos⁡(2𝜋𝑓𝑡)=
𝑚 𝑡 =
𝐵 (bandwidth)
𝐹 𝑐 =0
𝐵 (bandwidth)
𝐹 𝑐 =𝑓
𝐹 𝑐 =−𝑓
𝑢 𝑡 =𝑚 𝑡 cos 2𝜋𝑓𝑡 =

19. Down-conversion  bringing signal back to baseband
The receiver needs to perform an operation of down-conversion
The received signal is a high frequency signal in RF
Processing the data at these frequencies needs high clock digital circuits, which is impractical
We need to convert the data back to baseband and process the low frequency signals for decoding bits

20. Down-conversion  bringing signal back to baseband
𝑢 𝑡 =𝑚 𝑡 cos 2𝜋𝑓𝑡 =
𝐵 (bandwidth)
𝐹 𝑐 =−𝑓
𝐹 𝑐 =𝑓
Low pass filter eliminates this
Then, we recover baseband signal
𝐵 (bandwidth)
𝐹 𝑐 =2𝑓
𝐹 𝑐 =0
d 𝑡 =𝑢 𝑡 cos 2𝜋𝑓𝑡 = 𝑚 𝑡 cos 2 (2𝜋𝑓𝑡)
d 𝑡 =𝑚 𝑡 (1+𝑐𝑜𝑠 (4𝜋𝑓𝑡) ) 2
d 𝑡 = 𝑚(𝑡) 2 + 𝑐𝑜𝑠 (4𝜋𝑓𝑡) 2

21. Upconversion and Downconversion summary
𝑢(𝑡) 2 + 𝑐𝑜𝑠 (4𝜋𝑓𝑡) 2
m(t)
r(t) x x
cos⁡(2𝜋𝑓𝑡)
cos⁡(2𝜋𝑓𝑡)

22. Upconversion and Downconversion summary
𝑢(𝑡) 2 + 𝑐𝑜𝑠 (4𝜋𝑓𝑡) 2
I(t)
r(t) x x
cos⁡(2𝜋𝑓𝑡)
cos⁡(2𝜋𝑓𝑡)

23. Beyond amplitude modulation
We have learnt communication with amplitude modulation
There is a simple idea to double the data rate using QAM (quadrature amplitude modulation)

24. Quadrature amplitude modulation
Achieves double data rate compared to amplitude modulation alone x
cos⁡(2𝜋𝑓𝑡) x
cos⁡(2𝜋𝑓𝑡)
I(t) +
𝐼 𝑡 2 + 𝐼 𝑡 cos 4𝜋𝑓𝑡 + 𝑄 𝑡 𝑠𝑖𝑛 4𝜋𝑓𝑡 2
Q(t) x
𝑠𝑖𝑛⁡(2𝜋𝑓𝑡) x
sin⁡(2𝜋𝑓𝑡)
𝐼(𝑡)𝑐𝑜𝑠 2𝜋𝑓𝑡 +𝑄(𝑡)𝑠𝑖𝑛(2𝜋𝑓𝑡)
𝑄 𝑡 2 + 𝐼 𝑡 cos 4𝜋𝑓𝑡 + 𝑄 𝑡 𝑠𝑖𝑛 4𝜋𝑓𝑡 2

25. This scheme uses 16 symbols (4 bits per symbol), hence called 16 QAM
Symbols with QAM 𝐼
0010
0011
0001
0000
0110
0111
0100
1110
1111
1101
1100
1010
1011
1001
1000
3𝑉 4
𝑉 4
3𝑉 4
- 3𝑉 4
− 𝑉 4
𝑉 4 𝑄
− 𝑉 4
− 3𝑉 4
This scheme uses 16 symbols (4 bits per symbol), hence called 16 QAM

26. 64 QAM
Denser modulation can be used when symbol distortion is less in the channel

27. BPSK (binary phase shift keying)
Coarser modulation can be used when symbol distortion is huge

28. Channel multipath 𝐻 Tx Rx Channel Impulse Response Amplitude time
𝐹 𝑐 =𝑓
𝐹 𝑐 =−𝑓
Channel Frequency Response
𝐻 𝑓

29. Received data with channel multipath
𝑋 𝑓
𝐹 𝑐 =−𝑓
𝐹 𝑐 =𝑓
𝐹 𝑐 =𝑓
𝐹 𝑐 =−𝑓
𝐻 𝑓
𝐹 𝑐 =𝑓
𝐹 𝑐 =−𝑓
𝑌=𝐻∗𝑋
𝑌 𝑓 = 𝐻 𝑓 𝑋 𝑓

30. Deep channel fading 𝑋 𝑓 𝐻 𝑓 𝐹 𝑐 =−𝑓 𝐹 𝑐 =𝑓 𝐹 𝑐 =−𝑓 𝐹 𝑐 =𝑓 𝑌=𝐻∗𝑋
𝑌 𝑓 = 𝐻 𝑓 𝑋 𝑓

31. Revisit the transmitted spectrumF -F

32. Increase the symbol duration
Symbol rate R s =2∗𝐵 (bandwidth) F -F
Decreases the bandwidth of the signal -F F

33. Modified signal passed through the channel
𝑋 𝑓
𝐹 𝑐 =−𝑓
𝐹 𝑐 =𝑓
𝐻 𝑓
𝐹 𝑐 =−𝑓
𝐹 𝑐 =𝑓
𝐹 𝑐 =𝑓
𝐹 𝑐 =−𝑓
Data is unaffected since the fading frequencies do not overlap with data frequencies

34. Coping with fading
Decrease bandwidth so that data frequencies do not overlap with fading frequencies
This helped eliminate the effect of fading
Disadvantage: This would waste a lot of available bandwidth
Can we do better to achieve throughput proportional to the channel quality, without wasting any bandwidth

35. Coping with fading 𝑥 F -F
𝑥 1
𝑥 2
𝑥 3
𝑥 4 F -F
𝑥 1
𝑥 2
𝑥 3
𝑥 4

36. Coping with fading 𝑒 𝑖2𝜋 𝑓 1 𝑡 𝑒 𝑖2𝜋 𝑓 2 𝑡 𝑒 𝑖2𝜋 𝑓 3 𝑡 𝑒 𝑖2𝜋 𝑓 4 𝑡 x x𝑥 F -F
𝑓 1
𝑓 2
𝑓 3
𝑓 4
𝑥 1
𝑥 2
𝑥 3
𝑥 4
𝑥 1
𝑥 2
𝑥 3
𝑥 4 x
𝑥 1
𝑒 𝑖2𝜋 𝑓 1 𝑡 x
𝑥 2
𝑒 𝑖2𝜋 𝑓 2 𝑡 x
𝑥 3
𝑒 𝑖2𝜋 𝑓 3 𝑡 x
𝑥 4
𝑒 𝑖2𝜋 𝑓 4 𝑡

37. Coping with fading + + + 𝑒 𝑖2𝜋 𝑓 1 𝑡 𝑒 𝑖2𝜋 𝑓 2 𝑡 𝑒 𝑖2𝜋 𝑓 3 𝑡 x x x x F𝑥 𝑥
𝑥 1
𝑥 2
𝑥 3
𝑥 4 x +
𝑥 1
𝑒 𝑖2𝜋 𝑓 1 𝑡 x
𝑥 2
𝑥 1
𝑥 2
𝑥 3
𝑥 4
𝑒 𝑖2𝜋 𝑓 2 𝑡 x
𝑓 4
𝑓 3
𝑓 2
𝑓 1 +
𝑥 3
𝑒 𝑖2𝜋 𝑓 3 𝑡 + x
𝑥 4

38. + + + 𝑒 𝑖2𝜋 𝑓 1 𝑡 𝑒 𝑖2𝜋 𝑓 2 𝑡 𝑒 𝑖2𝜋 𝑓 3 𝑡 𝑒 𝑖2𝜋 𝑓 4 𝑡 x x x x F -F
𝑥(𝑡)
𝑥 1 (𝑡)
𝑥 1
𝑥 2
𝑥 3
𝑥 4
𝑒 𝑖2𝜋 𝑓 1 𝑡 + x
𝑥 2 (𝑡)
𝑒 𝑖2𝜋 𝑓 2 𝑡 +
𝑥 1
𝑥 2
𝑥 3
𝑥 4 x
𝑥 3 (𝑡)
𝑓 4
𝑓 3
𝑓 2
𝑓 1
𝑒 𝑖2𝜋 𝑓 3 𝑡 + x
𝑥 4 (𝑡)
𝑒 𝑖2𝜋 𝑓 4 𝑡

39. 𝑇𝑥=𝐼𝐹𝐹𝑇( 𝑥 1 𝑡 1 , 𝑥 2 (𝑡) …. , 𝑥 𝑖 (𝑡) … 𝑥 𝑛 (𝑡))F -F
𝑥 1
𝑥 2
𝑥 3
𝑥 4
𝑇𝑥=𝑥 1 (𝑡) 𝑒 𝑖2𝜋 𝑓 1 𝑡 + 𝑥 2 (𝑡) 𝑒 𝑖2𝜋 𝑓 2 𝑡 + 𝑥 3 (𝑡) 𝑒 𝑖2𝜋 𝑓 3 𝑡 + 𝑥 4 (𝑡) 𝑒 𝑖2𝜋 𝑓 4 𝑡
𝑇𝑥= ∑ 𝑥 𝑖 (𝑡)𝑒 𝑖2𝜋 𝑓 𝑖 𝑡
𝑓 4
𝑓 3
𝑓 2
𝑓 1
𝑇𝑥=𝐼𝐹𝐹𝑇( 𝑥 1 𝑡 1 , 𝑥 2 (𝑡) …. , 𝑥 𝑖 (𝑡) … 𝑥 𝑛 (𝑡))

40. OFDM (Orthogonal frequency division multiplexing) transmission
sin⁡(2𝜋𝑓𝑡) x
cos⁡(2𝜋𝑓𝑡) +
I = Real part
Q = Imag part
𝑓 1
𝑓 2
𝑓 3
𝑓 4
𝑓 5
IFFT
𝑓 4
𝑓 3
𝑓 2
𝑓 1
𝑓 5
𝑥 1
𝑥 2
𝑥 3
𝑥 4
𝑥 5
𝑥 1
𝑥 2
𝑥 3
𝑥 4
𝑥 5 −𝑓 𝑓

41. OFDM performance under deep fading
𝑋 𝑓
𝐹 𝑐 =−𝑓
𝐹 𝑐 =𝑓
𝐻 𝑓
𝐹 𝑐 =−𝑓
𝐹 𝑐 =𝑓
𝐹 𝑐 =𝑓
𝐹 𝑐 =−𝑓

42. OFDM reception x 𝑓 1 𝑓 2 𝑓 3 𝑓 4 𝑓 5 cos⁡(2𝜋𝑓𝑡) 𝑅𝑒 𝑚 = 𝐼(𝑡)
𝐼𝑚 𝑚 = 𝑄(𝑡)
𝑥 1
𝑓 1
𝐵𝑎𝑠𝑒𝑏𝑎𝑛𝑑
𝑓 4
𝑓 3
𝑓 2
𝑓 1
𝑓 5
𝑓 2
𝑥 2
𝑓 3
𝑥 3
𝐹𝐹𝑇
𝑓 4
𝑥 4
𝑓 5
𝑥 5
𝑠𝑖𝑛⁡(2𝜋𝑓𝑡)

43. OFDM vs Conventional Robust to deep fading
Very efficient, achieves capacity limits, used widely in LTE/WiFi
Robust to synchronization errors
Requires FFT/IFFT power intensive
High variation in signal amplitude – needs better h/w
Complete loss of performance under deep fading
Cannot reach maximum capacity
High synchronization overhead
Suitable for low power/battery- less communication
Low variation in signal amplitude